$ABCDE$ is a regular pentagon.  $AP$, $AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD$, $CB$ extended and $DE$ extended, respectively.  Let $O$ be the center of the pentagon.  If $OP = 1$, then find $AO + AQ + AR$.

[asy]

unitsize(2 cm);

pair A, B, C, D, E, O, P, Q, R;

A = dir(90);

B = dir(90 - 360/5);

C = dir(90 - 2*360/5);

D = dir(90 - 3*360/5);

E = dir(90 - 4*360/5);

O = (0,0);

P = (C + D)/2;

Q = (A + reflect(B,C)*(A))/2;

R = (A + reflect(D,E)*(A))/2;

draw((2*R - E)--D--C--(2*Q - B));

draw(A--P);

draw(A--Q);

draw(A--R);

draw(B--A--E);

label("$A$", A, N);

label("$B$", B, dir(0));

label("$C$", C, SE);

label("$D$", D, SW);

label("$E$", E, W);

dot("$O$", O, dir(0));

label("$P$", P, S);

label("$Q$", Q, dir(0));

label("$R$", R, W);

label("$1$", (O + P)/2, dir(0));

[/asy]
Solution: To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways.  First, we can divide regular pentagon $ABCDE$ into five congruent triangles.

[asy]

unitsize(2 cm);

pair A, B, C, D, E, O, P, Q, R;

A = dir(90);

B = dir(90 - 360/5);

C = dir(90 - 2*360/5);

D = dir(90 - 3*360/5);

E = dir(90 - 4*360/5);

O = (0,0);

P = (C + D)/2;

Q = (A + reflect(B,C)*(A))/2;

R = (A + reflect(D,E)*(A))/2;

draw((2*R - E)--D--C--(2*Q - B));

draw(A--P);

draw(A--Q);

draw(A--R);

draw(B--A--E);

draw((O--B),dashed);

draw((O--C),dashed);

draw((O--D),dashed);

draw((O--E),dashed);

label("$A$", A, N);

label("$B$", B, dir(0));

label("$C$", C, SE);

label("$D$", D, SW);

label("$E$", E, W);

dot("$O$", O, NE);

label("$P$", P, S);

label("$Q$", Q, dir(0));

label("$R$", R, W);

label("$1$", (O + P)/2, dir(0));

[/asy]

If $s$ is the side length of the regular pentagon, then each of the triangles $AOB$, $BOC$, $COD$, $DOE$, and $EOA$ has base $s$ and height 1, so the area of regular pentagon $ABCDE$ is $5s/2$.

Next, we divide regular pentagon $ABCDE$ into triangles $ABC$, $ACD$, and $ADE$.

[asy]

unitsize(2 cm);

pair A, B, C, D, E, O, P, Q, R;

A = dir(90);

B = dir(90 - 360/5);

C = dir(90 - 2*360/5);

D = dir(90 - 3*360/5);

E = dir(90 - 4*360/5);

O = (0,0);

P = (C + D)/2;

Q = (A + reflect(B,C)*(A))/2;

R = (A + reflect(D,E)*(A))/2;

draw((2*R - E)--D--C--(2*Q - B));

draw(A--P);

draw(A--Q);

draw(A--R);

draw(B--A--E);

draw(A--C,dashed);

draw(A--D,dashed);

label("$A$", A, N);

label("$B$", B, dir(0));

label("$C$", C, SE);

label("$D$", D, SW);

label("$E$", E, W);

dot("$O$", O, dir(0));

label("$P$", P, S);

label("$Q$", Q, dir(0));

label("$R$", R, W);

label("$1$", (O + P)/2, dir(0));

[/asy]

Triangle $ACD$ has base $s$ and height $AP = AO + 1$.  Triangle $ABC$ has base $s$ and height $AQ$.  Triangle $ADE$ has base $s$ and height $AR$.  Therefore, the area of regular pentagon $ABCDE$ is also \[\frac{s}{2} (AO + AQ + AR + 1).\]Hence, \[\frac{s}{2} (AO + AQ + AR + 1) = \frac{5s}{2},\]which means $AO + AQ + AR + 1 = 5$, or $AO + AQ + AR = \boxed{4}$.